Applications to Physics and Engineering

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 6.4 – Work
• Section 7.7 – Approximate Integration
• Section 8.3 – Applications to Physics and Engineering

Expected Educational Results

• Objective 14–1: Given the velocity of an object along a straight line, I can set up the integral to compute the displacement and total distance traveled of the object.
• Objective 14–2: Given the acceleration of an object along a straight line, I can set up the integral to compute the velocity, speed, displacement, and total distance traveled of the object.
• Objective 14–3: I can estimate a definite integral using a Riemann Sum.
• Objective 14–4: I can use the shape of a function to determine if a Riemann Sum is an over- or under-estimate of a definite integral.
• Objective 14–5: I can set up the integral to compute the moments and center of mass of a lamina.
• Objective 14–6: I can set up and evaluate the integral to compute the work on an object.
• Objective 14–7: I can set up and evaluate the integral using Hooke's Law to compute the work on a spring.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Applications to Physics and Engineering

Motion Along Line

Suppose the velocity, $v(t)$$v(t)$, of an object in motion along a line is known.

Definition: Speed

The speed of a particle in motion along a line is $\text{speed}=|v(t)|$$\text{speed}=\left|v(t)\right|$.

Definition: Displacement

The displacement of a particle in motion along a line is the net change in position. In other words, ${\displaystyle \text{displacement}={\int }_{t_1}^{t_2}v(t)dt}$$\displaystyle \text{displacement}=\int_{t_1}^{t_2}{v(t)}\,dt$.

Definition: Total Distance

The total distance of a particle in motion along a line is ${\displaystyle \text{distance}={\int }_{t_1}^{t_2}|v(t)|dt}$$\displaystyle \text{distance}=\int_{t_1}^{t_2}{\left|v(t)\right|}\,dt$.

Definition: Velocity

Suppose the acceleration, $a(t)$$a(t)$, of an object in motion along a line is known. The velocity of a particle in motion along a line is ${\displaystyle v={\int }_{t_1}^{t_2}a(t)dt}$$\displaystyle v=\int_{t_1}^{t_2}{a(t)}\,dt$.

Example 01: A particle is moving with velocity $v(t)=\cos (2\pi t)\frac{m}{s^2}$$v(t)=\cos{(2\pi t)} \frac{m}{s^2}$ at time $t$$t$.

1. Find the displacement of the particle from $t=0$$t=0$ to $t=\frac{5}{4}$$t=\frac{5}{4}$.
2. Find the total distance the particle traveled from $t=0$$t=0$ to $t=\frac{5}{4}$$t=\frac{5}{4}$.

Solution 01:

First, find the velocity at time $t$$t$:

${\displaystyle v(t)=\int a(t)dt=\int \cos (2\pi t)dt={\displaystyle \frac{\sin (2\pi t)}{2\pi }}+C}$$\displaystyle v(t)=\int{a(t)}\,dt =\int{\cos{(2\pi t)}\,dt} = \dfrac{\sin{(2 \pi t)}}{2\pi} +C$ in $\frac{m}{s}$$\frac{m}{s}$

Note that we can also compute the following, if needed:

The velocity at $t=3/4$$t=3/4$ seconds: ${\displaystyle v(3/4)=\frac{\sin (2\pi (3/4))}{2\pi }={\displaystyle \frac{\sin (2\pi (3/4))}{2\pi }}\frac{m}{s}}$$\displaystyle v(3/4) = \frac{\sin{(2 \pi (3/4))}}{2\pi} = \dfrac{\sin{(2 \pi (3/4))}}{2 \pi}\frac{m}{s}$

The speed at time $t$$t$: ${\displaystyle |v(t)|=|\frac{\sin (2\pi t)}{2\pi }+C|\frac{m}{s}}$$\displaystyle \left|v(t)\right| = \left|\frac{\sin{(2 \pi t)}}{2\pi}+C\right|\,\frac{m}{s}$

The speed at $t=3/4$$t=3/4$ seconds: ${\displaystyle |v(3/4)|=\left|\frac{\sin (2\pi (3/4))}{2\pi }\right|={\displaystyle \frac{|\sin (2\pi (3/4))|}{2\pi }}\frac{m}{s}}$$\displaystyle \left|v(3/4)\right| = \left|\frac{\sin{(2 \pi (3/4))}}{2\pi}\right| = \dfrac{|\sin{(2 \pi (3/4))}|}{2 \pi}\frac{m}{s}$

1. The displacement of the particle from $t=0$$t=0$ to $t=\frac{5}{4}$$t=\frac{5}{4}$.

${\displaystyle \text{displacement at time }t={\int }_0^{5/4}v(t)dt={\int }_0^{5/4}\frac{\sin (2\pi t)}{2\pi }dt}$$\displaystyle \text{displacement at time } t =\int_0^{5/4}{v(t)}\,dt =\int_0^{5/4}{\frac{\sin{(2\pi t)}}{2\pi}\,dt}$

${\displaystyle =-\frac{\cos (2\pi (5/4))}{4{\pi }^2}-(-\frac{\cos (2\pi (0))}{4{\pi }^2})=0-(-{\displaystyle \frac{1}{4{\pi }^2}})=\frac{1}{4{\pi }^2}m}$$\displaystyle = -\frac{\cos(2\pi(5/4))}{4\pi^2} - \left(-\frac{\cos(2\pi(0))}{4\pi^2}\right)= 0 - \left(-\dfrac{1}{4 \pi^2}\right)=\frac{1}{4\pi^2}\, m$

So, the particle is ${\displaystyle \frac{1}{4{\pi }^2}}m$$\dfrac{1}{4\pi^2}\, m$ from the starting point when the particle traveled from $t=0$$t=0$ to $t=5/4$$t=5/4$.

2. The total distance traveled by the particle from $t=0$$t=0$ to $t=\frac{5}{4}$$t=\frac{5}{4}$.

${\displaystyle \text{total distance traveled up to time }t={\int }_0^{5/4}|v(t)|dt={\int }_0^{5/4}\left|\frac{\sin (2\pi t)}{2\pi }\right|dt}$$\displaystyle\text{total distance traveled up to time } t = \int_0^{5/4}{\left|v(t)\right|}\,dt =\int_0^{5/4}{\left|\frac{\sin{(2\pi t)}}{2\pi}\right|\,dt}$

Consider the embedded DESMOS graph of $f(x)={\displaystyle \frac{\sin (2\pi t)}{2\pi }}$$f(x)=\dfrac{\sin{(2\pi t)}}{2\pi}$ (blue curve) and $g(x)=\left|{\displaystyle \frac{\sin (2\pi t)}{2\pi }}\right|$$g(x)=\left|\dfrac{\sin{(2\pi t)}}{2\pi}\right|$ (red curve):

To find the total area under $|v(t)|=\left|{\displaystyle \frac{\sin (2\pi t)}{2\pi }}\right|$$\left|v(t)\right| =\left| \dfrac{\sin{(2\pi t)}}{2\pi}\right|$ partition this curve and add up all the areas “under” the curve.

$\text{Total distance}=\text{Area under red portion of curve }+\text{ Area under blue portion }+\text{ Area under green portion}$$\text{Total distance} = \text{Area under red portion of curve } + \text{ Area under blue portion } + \text{ Area under green portion}$

${\displaystyle \Rightarrow \text{Total distance }={\int }_0^{0.5}{\displaystyle \frac{\sin (2\pi t)}{2\pi }}dx+{\int }_{0.5}^1{\displaystyle \frac{\sin (2\pi t)}{2\pi }}dx+{\int }_1^{5/4}{\displaystyle \frac{\sin (2\pi t)}{2\pi }}dx}$$\displaystyle\Rightarrow\text{Total distance } = \int_0^{0.5}{\dfrac{\sin{(2\pi t)}}{2\pi}\,dx} + \int_{0.5}^{1}{\dfrac{\sin{(2\pi t)}}{2\pi}\,dx} + \int_1^{5/4}{\dfrac{\sin{(2\pi t)}}{2\pi}\,dx}$

$={-{\displaystyle \frac{\cos (2\pi t)}{2\pi }}|}_0^{0.5}+{-{\displaystyle \frac{\cos (2\pi t)}{2\pi }}|}_{0.5}^1+{-{\displaystyle \frac{\cos (2\pi t)}{2\pi }}|}_1^{5/4}$$= \left.-\dfrac{\cos{(2 \pi t)}}{2\pi}\right|_0^{0.5} + \left.-\dfrac{\cos{(2 \pi t)}}{2\pi}\right|_{0.5}^{1} + \left.-\dfrac{\cos{(2 \pi t)}}{2\pi}\right|_1^{5/4}$

$={\displaystyle \frac{1}{2{\pi }^2}}+{\displaystyle \frac{1}{2{\pi }^2}}+{\displaystyle \frac{1}{4{\pi }^2}}={\displaystyle \frac{5}{4{\pi }^2}}$$= \dfrac{1}{2\pi^2} + \dfrac{1}{2\pi^2} + \dfrac{1}{4\pi^2} = \dfrac{5}{4\pi^2}$

So, the particle traveled a total distance of ${\displaystyle \frac{5}{4{\pi }^2}}m$$\dfrac{5}{4\pi^2}\, m$ from $t=0$$t=0$ to $t=5/4$$t=5/4$.

Investigation 01

A particle is moving along a line with constant acceleration of ${\displaystyle 4\frac{m}{s^2}}$$\displaystyle 4 \frac{m}{s^2}$. When $t=0$$t=0$, the velocity of the particle is ${\displaystyle -8\frac{m}{s}}$$\displaystyle -8 \frac{m}{s}$.

1. What is the velocity at time $t$$t$?
2. What is the velocity at $t=3$$t=3$?
3. What is the speed at time $t$$t$?
4. What is the speed at $t=3$$t=3$?
5. Find the displacement of the particle from $t=0$$t=0$ to $t=6$$t=6$.
6. Find the displacement of the particle from $t=4$$t=4$ to $t=8$$t=8$.
7. Find the total distance the particle traveled from $t=0$$t=0$ to $t=6$$t=6$.
8. Find the total distance the particle traveled from $t=4$$t=4$ to $t=8$$t=8$.

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Last Modified: Thursday, 15 October 2020 6:42 EDT